Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 175,... ISSN: 1072-6691. URL: or

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Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 175, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
THREE-POINT THIRD-ORDER PROBLEMS WITH A
SIGN-CHANGING NONLINEAR TERM
JOHNNY HENDERSON, NICKOLAI KOSMATOV
Abstract. In this article we study a well-known boundary value problem
u000 (t) = f (t, u(t)),
0
0 < t < 1,
00
u(0) = u (1/2) = u (1) = 0.
With u0 (η) = 0 in place of u0 (1/2) = 0, many authors studied the existence of
positive solutions of both the positone problems with η ≥ 1/2 and the semipositone problems for η > 1/2. It is well-known that the standard method
successfully applied to the semi-positone problem with η > 1/2 does not work
for η = 1/2 in the same setting. We treat the latter as a problem with a signchanging term rather than a semi-positone problem. We apply Krasnosel’skiı̆’s
fixed point theorem [4] to obtain positive solutions.
1. Introduction
We study the third-order nonlinear boundary-value problem
u000 (t) = f (t, u(t)),
0
0 < t < 1,
00
u(0) = u (1/2) = u (1) = 0.
(1.1)
(1.2)
with a sign-changing nonlinearity.
Equation (1.1) satisfying the three-point condition
u(0) = u0 (η) = u00 (1) = 0,
(1.3)
with η ≥ 1/2 has been studied by many authors [2, 7, 11]. We mention also relevant
results in [1, 3], where, under nonlocal conditions involving Stieltjes integrals, the
positone case was considered. A good theory of positive solutions for semi-positone
problems with η > 1/2 is developed in [5, 8, 9, 10] (and the references therein).
In particular, Yao [8] obtained a positive solution of the boundary value problem
similar to (1.1), (1.3). The author assumed that the function f : [0, 1] × R+ →
R satisfies the Carathéodory conditions and there exists a nonnegative function
h ∈ L1 [0, 1] such that f (t, u) ≥ h(t), (t, u) ∈ [0, 1] × R+ . Our paper is motivated
by [8] where, we believe, the idea of a non-constant lower bound −h(t) for the
inhomogeneous term was originally used for the boundary value problem similar
to (1.1), (1.2). The author refers to this type of problem as weakly semipositone.
2000 Mathematics Subject Classification. 34B15, 34B16, 34B18.
Key words and phrases. Green’s function; fixed point theorem; positive solutions;
third-order boundary-value problem.
c
2014
Texas State University - San Marcos.
Submitted June 10, 2014. Published August 14, 2014.
1
2
J. HENDERSON, N. KOSMATOV
EJDE-2014/175
Prior to [8], similar semipositone problems have been solved effectively [10] only for
f : [0, 1] × R → [−M, ∞) due to selection of h ≡ M > 0. As in [8], in this paper,
we need only f : [0, 1] × R+ → R.
Regardless of the choice of h, as a first step, one translates the semipositone
problem into a positone problem using the transformations
u 7→ v − u0
and f (·, u) 7→ f (·, v − u0 ) + h(·),
where u0 is a unique solution of the problem with the nonlinear term replaced with
h. Subsequently, the positone problem is converted into an integral equation, which
is shown to have one or, depending on conditions of f , several positive solutions.
Finally, an important feature of this approach is that it requires the inequality
v(t) ≥ u0 (t) to hold for a fixed point of the corresponding integral operator. This
comparison depends on the properties of Green’s function, or in particular, on the
function appearing in the definition of a cone, and the solution u0 . The case of
η = 1/2 stands alone since this type of approach used by many authors to study
the case η > 1/2 does not readily apply to the case η = 1/2. The difficulty arises
when we attempt to obtain the inequality v(t) ≥ u0 (t) for η = 1/2.
Since problem (1.1), (1.2) cannot be treated as a semipositone problem, we adopt
a new set of assumptions and consider a sign-changing nonlinearity. We are unaware
of any results on the case η = 1/2 with a sign-changing nonlinear term. Another
benefit is that we can also obtain new results for the case η > 1/2 with a signchanging nonlinearity by employing the concept of a sign-changing lower bound
g0 . We think that it would not be difficult to extend our results to the case of f
satisfying the Carathéodory conditions and even treat singularities as in [10]. Here
we settle for a continuous sign-changing nonlinear term.
2. Properties of Green’s function
Let g0 ∈ C[0, 1]. Then the differential equation
u000 (t) = g0 (t),
0 < t < 1,
(2.1)
satisfying the boundary condition (1.2) has a unique solution
Z
Z
t2 1
1 t
2
(t − s) g0 (s) ds −
g0 (s) ds
u0 (t) =
2 0
2 0
Z 1/2
1 Z 1
1
g0 (s) ds −
− s g0 (s) ds .
+t
2 0
2
0
Using Green’s function
1
1
1
G(t, s) = (t − s)2 χ[0,t] (s) + (t − t2 ) − t − s χ[0,1/2] (s),
2
2
2
for (t, s) ∈ [0, 1] × [0, 1], we have
Z 1
u0 (t) =
G(t, s)g0 (s) ds.
(2.2)
0
Let
s2
1
χ[0,1/2] (s) + χ[1/2,1] (s), s ∈ [0, 1].
2
8
We revisit the important properties [5] of (2.2) used in cone-theoretic methods:
G0 (s) = G(1/2, s) =
q(t)G0 (s) ≤ G(t, s) ≤ G0 (s),
(t, s) ∈ [0, 1] × [0, 1],
(2.3)
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THREE-POINT THIRD-ORDER PROBLEMS
3
where
q(t) = 4(t − t2 ).
(2.4)
Also,
Z
t∈[0,1]
1
G(t, s) ds =
L = max
0
1
,
12
(2.5)
and, for 0 < α < 1/2,
1−α
Z
1
(2 − 4α3 − 3α).
24
G0 (s) ds =
C=
α
(2.6)
Note that if
Z
1
g0 (t) dt ≥ 0,
t ∈ [0, 1],
(2.7)
t
then u0 (t) is concave in [0, 1]. If, in addition,
Z
Z 1/2
1 1
1
2
u0 (1) =
− s g0 (s) ds ≥ 0,
(1 − s) g0 (s) ds −
2 0
2
0
(2.8)
then u0 (t) ≥ 0. Note that neither (2.7) nor (2.8) requires g0 (t) ≥ 0 in all of [0, 1].
Moreover, if g0 (t) ≥ 0 in [0, 1] and g0 (t) > 0 in some [α, β] ⊂ [0, 1], then u0 (1) > 0.
This represents a difficulty due to the fact that one can not achieve the inequality
q(t) ≥ µu0 (t) in [0, 1] for any µ > 0 (as q(1) = 0 while u0 (1) > 0). For this reason,
the case η = 1/2 is forbidden in approaching (1.1), (1.3) as a semipositone problem.
If the identity takes place in (2.8), that is, u0 (1) = 0 is enforced, then we are in
position to compare q(t) and u0 (t) in the next lemma.
Lemma 2.1. Let g0 ∈ C[0, 1] satisfy (2.7) and suppose that the identity holds in
(2.8). Then there exists a constant µ > 0 such that
q(t) ≥ µu0 (t),
t ∈ [0, 1].
(2.9)
Proof. Since the function q − µu0 vanishes at the end-points of [0, 1], it suffices to
obtain µ > 0 such that −q 00 (t) ≥ −µu000 (t) in [0, 1]. That is,
Z 1
8≥µ
g0 (s) ds, t ∈ [0, 1].
t
By (2.7), there exists 0 < τ < 1 and µ > 0 such that
Z 1
Z 1
µ
g0 (s) ds ≤ µ
g0 (s) ds = 8.
t
(2.10)
τ
Suppose that the function f in (1.1) satisfies
(H1) f ∈ C([0, 1] × R+ , R);
(H2) there exists a function g0 ∈ C[0, 1] such that
(a) f (t, z) + g0 (t) ≥ 0 in [0, 1] × R+ ;
R1
(b) for all t ∈ [0, 1], t g0 (s) ds ≥ 0;
(c)
Z
Z 1/2
1 1
1
2
(1 − s) g0 (s) ds −
− s g0 (s) ds = 0.
2 0
2
0
4
J. HENDERSON, N. KOSMATOV
EJDE-2014/175
Remark 2.2. It is easy to find a function g0 satisfying (H2) (b) and (c). For
example, one can take g0 (t) = a(2t − 1), a > 0. Of course, an example of f (t, z)
that fits (H2) (b) is also easy to obtain.
Remark 2.3. If the inequality (2.8) is replaced with the strict inequality, we cannot
expect Lemma 2.1 to hold. So, in this paper, we need the identity in (H2) (c). If,
instead of u0 (1/2) = 0, we impose u0 (η) = 0 with η > 1/2, then the problem (2.1),
(1.3) has a unique solution
Z
Z
Z η
Z 1
1 t
t2 1
u0 (t) =
(t−s)2 g0 (s) ds−
g0 (s) ds+t η
g0 (s) ds−
(η −s)g0 (s) ds .
2 0
2 0
0
0
Again, the assumption (H1) (b) guarantees that u0 is concave in [0, 1]. So, if
Z
Z
Z η
1 1
1 1
2
(1 − s) g0 (s) ds + η −
g0 (s) ds −
(η − s)g0 (s) ds ≥ 0,
u0 (1) =
2 0
2 0
0
then u(t) ≥ 0 in [0, 1]. Similarly, the analogue of q(t), in this case [5], is
1
p(t) = 2 (2ηt − t2 ).
η
Noting that p(1) 6= 0 and u0 , p are concave concave in [0, 1], we can easily obtain
an analogue of Lemma 2.1 asserting the existence of µ > 0 such that p(t) ≥ µu0 (t)
in [0, 1]. This would give a more general result than in [10, Lemma 2.1 (4)], which
is derived for g0 ≡ M > 0. This would also allow us to extend the results of [8]
concerning an analogue of (1.1), (1.3), where h(t), which serves the purpose of g0 (t),
is assumed to be nonnegative.
We modify the problem (1.1), (1.2) as follows. First, we define
(
f (t, z) + g0 (t), (t, z) ∈ [0, 1] × [0, ∞),
fp (t, z) =
f (t, 0) + g0 (t), (t, z) ∈ [0, 1] × (−∞, 0).
Next, we consider the equation
v 000 (t) = fp (t, v(t) − u0 (t)),
t ∈ (0, 1),
(2.11)
under the boundary conditions (1.2). We can easily obtain the next lemma.
Lemma 2.4. The function u is a positive solution of the boundary value problem
(1.1), (1.2) if, and only if, the function v = u + u0 is a solution of the boundary
value problem (2.11), (1.2) satisfying v(t) ≥ u0 (t) in [0, 1].
In the Banach space B = C[0, 1] endowed with usual max-norm, we consider the
operator
Z 1
T v(t) =
G(t, s)fp (s, v(s) − u0 (s)) ds,
(2.12)
0
where G(t, s) is given by (2.2). By (H1), T : B → B is completely continuous.
Using the function q defined by (2.4), we introduce the cone
C = {v ∈ B : v(t) ≥ q(t)kvk, t ∈ [0, 1]}.
By (2.3), T : C → C and it is also easy to show that a fixed point of T is a solution
of (2.11), (1.2). In particular,
v(t) ≥ γkvk,
2
t ∈ [τ, 1 − τ ],
(2.13)
where γ = mint∈[α,1−α] q(t) = 4(α − α ), and κ = maxt∈[α,1−α] q(t) = q(1/2) = 1.
EJDE-2014/175
THREE-POINT THIRD-ORDER PROBLEMS
5
Theorem 2.5 ([4]). Let B be a Banach space and let C ⊂ B be a cone in B. Assume
that Ω1 , Ω2 are open with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let
T : C ∩ (Ω2 \ Ω1 ) → C
be a completely continuous operator such that either
(i) kT uk ≤ kuk, u ∈ C ∩ ∂Ω1 , and kT uk ≥ kuk, u ∈ C ∩ ∂Ω2 , or;
(ii) kT uk ≥ kuk, u ∈ C ∩ ∂Ω1 , and kT uk ≤ kuk, u ∈ C ∩ ∂Ω2 .
Then T has a fixed point in C ∩ (Ω2 \ Ω1 ).
3. Positive solutions
To use Theorem 2.5, following [8] we introduce the “height” functions φ, ψ :
R+ → R+ defined by
φ(r) = max{f (t, z − u0 (t)) + g0 (t) : t ∈ [0, 1], z ∈ [0, r]},
ψ(r) = min{f (t, z − u0 (t)) + g0 (t) : t ∈ [α, 1 − α], z ∈ [γr, r]}.
Now we present our main results.
Theorem 3.1. Assume that (H1) and (H2) hold. Suppose that there exist r, R > 0
such that µ1 < r < R, where µ > 0 satisfies (2.9), (2.10), and
(H3) φ(r) ≤ 12r and ψ(R) ≥
24R
2−4α3 −3α .
Then the boundary-value problem (1.1), (1.2) has at least one positive solution.
Proof. Let
Ω1 = {v ∈ B : kvk < r},
Ω2 = {v ∈ B : kvk < R}.
For u ∈ C ∩ ∂Ω1 , we have v(s) − u0 (s) ≥ q(s)kvk − u0 (s) ≥ (µr − 1)u0 (s) ≥ 0,
s ∈ [0, 1]. This implies that
fp (s, v(s) − u0 (s)) = f (s, v(s) − u0 (s)) + g0 (s),
s ∈ [0, 1].
In particular,
f (s, v(s) − u0 (s)) + g0 (s) ≤ φ(r),
s ∈ [0, 1], 0 ≤ v(s) ≤ r.
Thus, by (2.5) and (H3),
Z
1
kT vk = max
t∈[0,1]
G(t, s)fp (s, v(s) − u0 (s)) ds
0
Z
1
≤ max
t∈[0,1]
G(t, s) ds φ(r)
0
1
φ(r) ≤ r.
12
That is, kT vk ≤ kvk for all v ∈ C ∩ ∂Ω1 .
Let v ∈ C ∩ ∂Ω2 . Since R > r, we have v(s) − u0 (s) ≥ (µR − 1)u0 (s) ≥ 0,
s ∈ [0, 1]. Then, for all s ∈ [α, 1 − α], we have, recalling (2.13),
= Lφ(r) =
R ≥ v(s) ≥ q(s)kvk ≥ γR.
Hence
fp (s, v(s) − u0 (s)) = f (s, v(s) − u0 (s)) + g0 (s) ≥ ψ(R),
6
J. HENDERSON, N. KOSMATOV
EJDE-2014/175
for s ∈ [α, 1 − α], γR ≤ v(s) ≤ R. Then, by (2.6) and (H3),
Z 1
G(t, s)fp (s, v(s) − u0 (s)) ds
kT vk = max
t∈[0,1]
0
Z
1−α
G(t, s)fp (s, v(s) − u0 (s)) ds
≥ max
t∈[0,1]
α
Z 1−α
≥ max
t∈[0,1]
q(t)G0 (s) ds ψ(R)
α
Z
1−α
= max q(t)
t∈[0,1]
G0 (s) ds ψ(R)
α
= κCψ(R) ≥ R.
That is, kT vk ≥ kvk for all v ∈ C ∩ ∂Ω2 .
By Theorem 2.5, there exists v0 ∈ C with u(t) = v0 (t)−u0 (t) ≥ (µr −1)u0 (t) ≥ 0
in [0, 1]. By Lemma 2.4, u is a positive solution of the sign-changing problem (1.1),
(1.2).
Now we give an example of the right side of (1.1) satisfying the assumptions of
Theorem 3.1.
Example. Let f (t, z) = 6z 2 +32(1−2t) for z ≥ 0, t ∈ [0, 1]. Then f (t, z)+g0 (t) ≥ 0
with g0 (t) = 32(2t − 1). Of course, (H1) and (H2) hold and
Z 1
Z 1
g0 (t) ds ≤
g0 (t) ds = 8.
t
1/2
Hence we can choose µ = 1 and note that µr > 1, if we choose r = 2. Then,
recalling that v(s) − u0 (s) ≥ 0,
f (t, v(s) − u0 (s)) + g0 (t) = 6(v(s) − u0 (s))2 ≤ 6kvk2 = 24 = 12r,
for all v ∈ C ∩ ∂Ω1 . This shows that the first condition of (H3) is fulfilled.
It is easy to see that
Z 1
Z 1
8
ku0 k ≤ max
G(t, s)|g0 (s)| ds ≤ max
G(t, s) ds kg0 k = .
3
t∈[0,1] 0
t∈[0,1] 0
Let now α = 1/4 so that C = 19/384 and γ = 4(α − α2 ) = 3/4. Then, for all
s ∈ [1/4, 3/4], v ∈ C ∩ ∂Ω2 , where R = 13, we have
f (t, v(s) − u0 (s)) + g0 (t) = 6(v(s) − u0 (s))2
≥ 6(γkvk − ku0 k)2
7225
39 8 2
−
=
=6
4
3
24
4992
24R
>
=
.
19
2 − 4α3 − 3α
The above shows that the second part of (H3 ) is also verified. Hence a solution v0
exists in the cone and 2 ≤ kv0 k ≤ 13.
The next result can be shown along the similar lines.
Theorem 3.2. Assume that (H1) and (H2) hold. Suppose that there exist r, R > 0
such that µ1 < r < R, where µ > 0 satisfies (2.9), (2.10), and
EJDE-2014/175
THREE-POINT THIRD-ORDER PROBLEMS
(H4) φ(R) ≤ 12R and ψ(r) ≥
7
24r
2−4α3 −3α .
Then the boundary0value problem (1.1), (1.2) has at least one positive solution.
In conclusion of this paper presents a multiplicity result for (1.1), (1.2) which
now is considered as a nonlinear eigenvalue problem. That is,
u000 (t) = λf (t, u(t)),
0 < t < 1,
(3.1)
subject to (1.2). The result including the assumptions and the method of proof
echoes that of Ma [6], where a fourth order semipositone boundary-value problem
with dependence on the first derivative was studied. The presence of the parameter
λ > 0 provides an additional control on the growth of the right side. We introduce
a new set of assumptions as follows:
(M1) there exists an interval [α, 1 − α] ⊂ (0, 1) such that
lim
u→∞
f (t, u)
= ∞,
u
uniformly in [α, 1 − α];
(M2) f (t, 0) > 0, t ∈ [0, 1].
Our next result is a multiplicity criterion.
Theorem 3.3. Assume that (H1), (H2), (M1), (M2) hold. Then the boundaryvalue problem (3.1), (1.2) has at least two positive solutions provided λ > 0 is small
enough.
Proof. We will construct open nonempty subsets Ωi = {v ∈ C : kvk = Ri }, i =
1, . . . , 4. Now, we consider the operator
Z 1
T v(t) = λ
G(t, s)fp (s, v(s) − λu0 (s)) ds,
0
where u0 is the solution of u000 = g0 subject to (1.2) and fp as above. Let the
R1 > 0. Then
Z 1
kT vk = max λ
G(t, s)fp (s, v(s) − λu0 (s)) ds ≤ λLφ(R1 ) ≤ R1
t∈[0,1]
0
for all v ∈ C ∩ ∂Ω1 , provided
Lφ(R1 )
.
R1
Let v ∈ C ∩ ∂Ω2 , where R2 > R1 . Then, by Lemma 2.1 with
Z 1
µ max
g0 (s) ds = 8.
λ≤
t∈[0,1]
(3.2)
t
Note that the equation in (M2) holds with fp in place of f . Thus given A > 0,
there exists h ≥ γ2 R2 such that fp (t, z) > Az for all z ≥ h and t ∈ [α, 1 − α]. For
every λ in (3.2), there exists a constant A > 0 such that
1
λCγA ≥ 1,
2
where C is given by by (2.6). For all s ∈ [α, 1 − α], we have
v(s) − λu0 (s) ≥ v(s) −
λ
λ
1
γ
q(s) = v(s) −
v(s) ≥ v(s) ≥ R2
µ
µR2
2
2
(3.3)
8
J. HENDERSON, N. KOSMATOV
EJDE-2014/175
provided
λ≤
µR2
.
2
(3.4)
Hence
fp (s, v(s) − λu0 (s)) ≥ A(v(s) − λu0 (s)) ≥
γA
R2 ,
2
s ∈ [α, 1 − α].
Then, by (3.3)), and recalling that κ = 1,
Z 1
kT vk = max λ
G(t, s)fp (s, v(s) − λu0 (s)) ds
t∈[0,1]
0
Z
1−α
≥ λ max
t∈[0,1]
q(t)G0 (s) ds
α
Z
1−α
G0 (s) ds
= λ max q(t)
t∈[0,1]
α
γA
R2
2
γA
R2
2
γA
= λκC
R2 ≥ R2 .
2
That is, kT vk ≥ kvk for all v ∈ C ∩ ∂Ω2 . As in Theorem 3.1, we have a solution v1
such that R1 ≤ kv1 k ≤ R2 for every
0 < λ ≤ λ0 = min
µR2 R1
,
.
Lφ(R1 ) 2
To make use of the assumption (M2 ), we note that there exist a, b > 0 such that
f (t, z) ≥ b for all t ∈ [0, 1] and z ∈ [0, a] and introduce a “truncation” of f given by
(
f (t, z), (t, z) ∈ [0, 1] × [0, a]),
ft (t, z) =
f (t, a), (t, z) ∈ [0, 1] × (a, ∞).
Consider now
u000 (t) = λft (t, u(t)),
0 < t < 1,
(3.5)
subject to (1.2). The operator, whose fixed point will be shown to be (a second)
solution of (1.1), (1.2), is
Z 1
T v(s) = λ
G(t, s)ft (s, v(s)) ds.
0
Choose R3 < min{R1 , a}. Then, as in the proof of Theorem 3.1,
kT vk ≤ λLφ(R3 ),
where φ(R3 ) = max{f (t, z) : t ∈ [0, 1], z ∈ [0, R3 ]}. So, if
λ < min
R3
, λ0 ,
Lφ(R3 )
then kT vk ≤ kvk for all v ∈ C ∩ ∂Ω3 . Choose λ according to (3.6). Since
lim
z→0+
b
ft (t, z)
≥ lim
=∞
+
z
z→0 z
uniformly in [0, 1]. Hence there exists 0 < R4 < R3 such that
ft (t, z) ≥ Bz,
t ∈ [0, 1], z ∈ [0, R4 ],
(3.6)
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THREE-POINT THIRD-ORDER PROBLEMS
9
where
Z
λBD ≥ 1,
t∈[0,1]
1
G(t, s)q(s) ds.
D = max
0
Then, for all v ∈ C ∩ ∂Ω4 ,
1
Z
kT vk = max λ
t∈[0,1]
G(t, s)ft (s, v(s)) ds
0
Z
1
≥ max λB
t∈[0,1]
G(t, s)v(s) ds
0
Z
≥ λB max
t∈[0,1]
1
G(t, s)q(s)R4 ds
0
= λBDR4
≥ kvk.
Thus, there exists a positive solution v2 with R4 ≤ kv2 k ≤ R3 for every λ > 0
satisfying (3.6). Finally, since R3 < R1 , the solutions are distinct.
Acknowledgments. The article was prepared while the second author was on
sabbatical leave at Baylor University. He is grateful to Baylor University for its
support and hospitality.
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Johnny Henderson
Department of Mathematics, Baylor University, Waco, TX 76798-7328, USA
E-mail address: Johnny [email protected]
10
J. HENDERSON, N. KOSMATOV
EJDE-2014/175
Nickolai Kosmatov
Department of Mathematics and Statistics, University of Arkansas at Little Rock,
Little Rock, AR 72204-1099, USA
E-mail address: [email protected]
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